Monday, June 17, 2019

probability - finding expected value using CDF



If it is known that the random variable X has a CDF:



FX(t)={0t<1121t<034+t20t<12112t



I wish to find E[(Xa)2] for all aR




I found that E[(Xa)2]=E[X2]2aE[X]+a2



my teacher solved it this way:
enter image description here



what I'm trying to understand is the reason for the integrals used to find E[X],E[X2]. Why does this work?


Answer



The first equality in the computation of E[X2] follows from the fact that, for a non-negative random variable Y with CDF G, we have that




E[Y]=0ydG(y)=0(1G(t))dt.



You can find a proof of this here.



To see the first equality for E[X], we use the above result, along with the fact for a non-positive random variable Y with CDF G,



E[Y]=0ydG(y)=0G(t)dt.



You can prove this in a way similar to the analogous result for non-negative random variables.




Putting the two facts together allows us to write



E[X]=xdFX(x)=0xdFX(x)+0xdFX(x)=0(1FX(t))dt0FX(t)dt.


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